Question

A student is able to see printing in a book clearly when the book is 40cm away,
but not when it is closer. Determine the focal length suitable for the correction
of the defect.

Answer 1

Hypermetropia is a condition of the eye where the person is not able to see things clearly when nearer to the eye. The normal near point of the eye is 25 cm .

NUMERICAL CALCULATION

1/focal length (f) = 1/ object distance (u) + 1/ image distance (v)

Since the image formed is virtual i.e -ve

1/f = 1/u - 1/v

1/f = 1/25 - 1/40

1/f = 40/1000 - 25/1000

1/f = 15/1000

f = 1000/15

f ≈ 66.67 cm

Focal Length suitable for the correction of the defect is 66.67 cm

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Answer 2

Answer:

A student is able to see printing in a book clearly when the book is 40cm away, but not when it is closer. Determine the focal length suitable for the correction of the defect.

Explanation:

• Here is student is able to see the printing in a book clearly but he is not able to see the objects which are closer to him, That is he is not able to see the objects clearly which are close to him.
• Thus he have the problem of hypermetropia.
• Hypermetropia is the problem or the condition were the person is not able to see the object which are very near to him that is which are closer to the eyes. The normal point of the eye is 25 cm.
• So here we have to calculate the focal length which is suitable for the student to correct the defect.
• As we know the formula of focal length-
• 1/ Focal length (f)= 1/ object distance+ 1/ image distance (v).
• Here the image formed is virtual, therefore it is negative.
• $\frac{1}{f}= \frac{1}{u} -\frac{1}{v} \\\frac{1}{f}= \frac{1}{25} -\frac{1}{40} \\\frac{1}{f} =\frac{40}{1000}- \frac{25}{1000}\\ \frac{1}{f} =\frac{15}{1000}\\ f= \frac{1000}{15}\\ f= 66.67 cm$
• Hence after solving the formula for focal length we concluded that the focal length is 66.67 cm respectively.